POJ 2653 Pick-up sticks(线段相交)

题意：给定n个木棍依次放下，要求最终判断没被覆盖的木棍是哪些。

思路:快速排斥以及跨立实验可以判断线段相交。

 1 #include<algorithm>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<iostream>
 6 const double eps=1e-10;
 7 struct Point{
 8     double x,y;
 9     Point(){}
10     Point(double x0,double y0):x(x0),y(y0){}
11 }s[200005],e[200005];
12 int ans[200005],st[200005],n;
13 double operator *(Point p1,Point p2){
14     return p1.x*p2.y-p1.y*p2.x;
15 }
16 Point operator -(Point p1,Point p2){
17     return Point(p1.x-p2.x,p1.y-p2.y);
18 }
19 bool judge(Point p1,Point p2,Point p3,Point p4){
20     if (std::min(p1.x,p2.x)>std::max(p3.x,p4.x)||
21         std::min(p1.y,p2.y)>std::max(p3.y,p4.y)||
22         std::min(p3.x,p4.x)>std::max(p1.x,p2.x)||
23         std::min(p3.y,p4.y)>std::max(p1.y,p2.y))
24     return 0;//快速排斥实验
25     double a,b,c,d;
26     a=(p2-p1)*(p3-p1);
27     b=(p2-p1)*(p4-p1);
28     c=(p4-p3)*(p1-p3);
29     d=(p4-p3)*(p2-p3);
30     return (a*b<=eps)&&(c*d<=eps);//跨立实验
31 }
32 int main(){
33     while (scanf("%d",&n)!=EOF&&n){
34         for (int i=1;i<=n;i++) ans[i]=0;
35         for (int i=1;i<=n;i++){
36             scanf("%lf%lf%lf%lf",&s[i].x,&s[i].y,&e[i].x,&e[i].y);
37         }
38         for (int i=1;i<=n;i++)
39          for (int j=i+1;j<=n;j++)
40           if (judge(s[i],e[i],s[j],e[j]))
41          { ans[i]=1;
42            break;
43          }
44         int top=0;
45         for (int i=1;i<=n;i++)
46          if (!ans[i])
47           st[++top]=i;
48         printf("Top sticks:");
49         for (int i=1;i<top;i++)
50          printf(" %d,",st[i]);
51         printf(" %d.\n",st[top]);
52     }
53 }